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Prove this for every $n>1$ (belongs to $\mathbb{N}$ )

$$\displaystyle \int_{0}^{1}\left( \frac{x^{2n+3} - x^{2n+1}}{1+x} \right) \, \mathrm{d}x =\frac{1}{2n+3} - \frac{1}{2n+2}$$

I don't see how one can obtain this.

thanks answered ...

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Factor $x^{2n+1}$ from the numerator of the integrand. Then factor completely. It should be obvious what to do afterwards. – David Mitra Feb 19 '14 at 23:19

1 Answer 1

Let's factor the top:

$x^{2n+3} = x^{2n+1}x^2$ Then, the numerator becomes, $x^{2n+1}(x^2-1) = x^{2n+1}(x+1)(x-1)$

Now the integration should be straightforward.

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Nice one! ${}{}{}$ – Daniel Montealegre Feb 28 '14 at 1:10

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